Question 44: Gravitation

Question

The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $\frac{g}{9}$, where $g$ is acceleration due to gravity on earth's surface, is
$\_\_\_\_$ .

Accepted Answer

For a point at height $h$ above the surface of the earth, acceleration due to gravity is
$$ g_h = g\left(\frac{R}{R+h}\right)^2 $$
Given that
$$ g_h = \frac{g}{9} $$
So,
$$ g\left(\frac{R}{R+h}\right)^2 = \frac{g}{9} $$
Cancel $g$ from both sides:
$$ \left(\frac{R}{R+h}\right)^2 = \frac{1}{9} $$
Taking square root on both sides,
$$ \frac{R}{R+h} = \frac{1}{3} $$
So,
$$ 3R = R + h $$
$$ h = 2R $$
Hence, the correct answer is:
$$ \boxed{2R} $$
So, Option C is correct.

Answer

$\sqrt{3} R$

Answer

$2 \sqrt{2} R$

Answer

$2 R$

Answer

${\frac{4}{9} R}$

Gravitation

The height in terms of radius of the earth (R), at which the acceleration due to gravity becomes \frac{g}{9}, where g is acceleration due to gravity on earth's surface, is \_\_\_\_ .

Choose the correct answer: